The half life of a radioactive substance is 99738 seconds. How log will it take until there is 6.9% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{99738}$ . This gives the equation $LaTeX: \displaystyle 0.069 = e^{-\frac{\ln(2)}{99738}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.069)= \frac{-t\ln(2)}{99738}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 99738\ln(0.069) }{ \ln(2) }$ . It will take about about 384715.4 seconds.